Question 388048
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ h\right)^2\ +\ \left(y\ -\ k\right)^2\ =\ r^2]


is the equation of a circle centered at *[tex \Large \left(h,\,k\right)] with radius *[tex \Large r]


Take your standard form equation and complete the square on each of the variables.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 12x\ +\ 6y\ +\ 29\ =\ 0]


First move the constant to the RHS and group like variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 12x\ +\ y^2\ +\ 6y\ =\ -\ 29]


Take the coefficient on the first degree *[tex \Large x] term, divide by two, square the result, and add the squared result to both sides.  Do the same thing with the *[tex \Large y] variable.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 12x\ +\ 36\ +\ y^2\ +\ 6y\ +\ 9\ =\ -\ 29\ +\ 36\ +\ 9]


Factor the two perfect square trinomials and collect terms in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (-6)\right)^2\ +\ \left(y\ -\ (-3)\right)^2\ =\ 16]


From which you can determine the center by inspection and the radius by taking the positive square root of the RHS.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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